Question

. A 5 kg block is suspended from a spring having a stiffness of 300 N/m. If the block is acted upon by a vertical force F = 17 sin(8t) N, where t is in seconds, determine the equation which describes the motion of the block when it is pulled down 100 mm from the equilibrium position and released from rest at t = 0.Assume that positive displacement is downward.

Answer 1

The equation that describes the motion of the block can be found using Newton's second law and Hooke's law. When the block is pulled down 100 mm from the equilibrium position and released from rest at t=0, the equation can be written as 5 * x''(t) = -300 * x(t) + 17 * sin(8t).

Step-by-step explanation:

The equation that describes the motion of the block can be found using Newton's second law and Hooke's law.

First, let's find the equilibrium position. When the block is pulled down 100 mm from the equilibrium position, we can convert the distance to meters by dividing it by 1000. So, the distance from the equilibrium position is 0.1 m.

Next, we can use Newton's second law: F = ma. The vertical force acting on the block is given by F = 17 sin(8t) N. The mass of the block is 5 kg. The acceleration of the block is the second derivative of its position with respect to time, which can be denoted as x''(t).

Using Hooke's law, we know that the force exerted by the spring can be given by F = -kx(t). Here, x(t) is the displacement of the block from the equilibrium position and k is the stiffness of the spring, which is given as 300 N/m.

Combining these equations, we can write:

5 * x''(t) = -300 * x(t) + 17 * sin(8t)

Answer 2

x(t) = 0.053 sin(8t) + 0.425t + 0.1

Step-by-step explanation:

Using Newton’s 2nd law we can calculate the acceleration equation of the block with respect to time t

`a(t) = F/m = 17 sin(8t) / 5 = 3.4 sin(8t)`

From here we can derive the velocity equation by integrating the acceleration equation

`v(t) = \int\limits {a(t)} \, dt =- 3.4 cos(8t)/8 + v_0 = -0.425 cos(8t) + C_v`

Since the block is released from rest at t = 0, it means the velocity is 0 at t = 0

`v(0) = 0`

`-0.425cos(0) + C_v = 0`

`C_v = 0.425 m/s`

`v(t) = -0.425 cos(8t) + 0.425`

To find the equation of motion, we integrate the velocity equation:

`x(t) = \int\limit {v(t)} \, dt = -0.425sin(8t)/8 + 0.425t + C_x`

`x(t) = 0.053 sin(8t) + 0.425t + C_x`

At t = 0, object is released at position x = 100 mm = 0.1 m:

`x(0) = 0.1`

`0.053 sin(0) + 0.425*0 + C_x = 0.1`

`C_x = 0.1`

`x(t) = 0.053 sin(8t) + 0.425t + 0.1`